Volume 2009, Article ID 624631,30pages doi:10.1155/2009/624631
Research Article
Complementary Lidstone Interpolation and
Boundary Value Problems
Ravi P. Agarwal,
1, 2Sandra Pinelas,
3and Patricia J. Y. Wong
41Department of Mathematical Sciences, Florida Institute of Technology, Melbourne, FL 32901, USA 2Mathematics and Statistics Department, King Fahd University of Petroleum and Minerals,
Dhahran 31261, Saudi Arabia
3Department of Mathematics, Azores University, R. M˜ae de Deus, 9500-321 Ponta Delgada, Portugal 4School of ELectrical & Electronic Engineering, Nanyang Technological University, Singapore 639798
Correspondence should be addressed to Ravi P. Agarwal,[email protected]
Received 21 August 2009; Revised 5 November 2009; Accepted 6 November 2009
Recommended by Donal O’Regan
We shall introduce and construct explicitly the complementary Lidstone interpolating polynomial
P2mtof degree 2m, which involves interpolating data at the odd-order derivatives. ForP2mt we will provide explicit representation of the error function, best possible error inequalities, best possible criterion for the convergence of complementary Lidstone series, and a quadrature formula with best possible error bound. Then, these results will be used to establish existence and uniqueness criteria, and the convergence of Picard’s, approximate Picard’s, quasilinearization, and approximate quasilinearization iterative methods for the complementary Lidstone boundary value problems which consist of a2m1th order differential equation and the complementary Lidstone boundary conditions.
Copyrightq2009 Ravi P. Agarwal et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
In our earlier work1,2we have shown that the interpolating polynomial theory and the qualitative as well as quantitative study of boundary value problems such as existence and uniqueness of solutions, and convergence of various iterative methods are directly connected. In this paper we will extend this technique to the followingcomplementary Lidstone boundary value probleminvolving an odd order differential equation
−1mx2m1t ft,xt, t∈0,1, m≥1, 1.1
and the boundary data at the odd order derivatives
Here x x, x, . . . , xq, 0 ≤ q ≤ 2mbut fixed, and f : 0,1×Rq1 → R is continuous at least in the interior of the domain of interest. Problem1.1,1.2complementsLidstone boundary value problemnomenclature comes from the expansion introduced by Lidstone3
in 1929, and thoroughly characterized in terms of completely continuous functions in the works of Boas 4, Poritsky 5, Schoenberg6–8, Whittaker9,10, Widder11,12, and otherswhich consists of an even-order differential equation and the boundary data at the even-order derivatives
−1mx2mt ft,xt, t∈0,1, m≥1, x2i0 ai, x2i1 bi, 0≤i≤m−1.
1.3
Problem1.3has been a subject matter of numerous studies in the recent years13–45, and others.
InSection 2, we will show that for a given functionx : C2m10,1 → Rexplicit representations of the interpolation polynomialP2mtof degree 2msatisfying the conditions
P2m0 x0, P2m2i−10 x2i−10, P2m2i−11 x2i−11, 1≤i≤m 1.4
and the corresponding residue termRt xt−P2mtcan be deduced rather easily from
our earlier work on Lidstone polynomials46–48. Our method will avoid unnecessarily long procedure followed in49to obtain the same representations ofP2mtandRt.We will also
obtain error inequalities
xkt−P2mkt≤C2m1,kmax
0≤t≤1
x2m1t, k0,1, . . . ,2m, 1.5
where the constantsC2m1,k are the best possible in the sense that in1.5equalities hold if and only ifxtis a certain polynomial. The best possible constantC2m1,0was also obtained in49; whereas they left the cases 1≤k≤2mwithout any mention. InSection 2, we will also provide best possible criterion for the convergence of complementary Lidstone series, and a quadrature formula with best possible error bound.
If f 0 then the complementary Lidstone boundary value problem 1.1, 1.2
obviously has a unique solution xt P2mt; iff is linear, that is,f qi0aitxi then
1.1,1.2gives the possibility of interpolation by the solutions of the differential equation
1.1. In Sections 3–5, we will use inequalities 1.5 to establish existence and uniqueness criteria, and the convergence of Picard’s, approximate Picard’s, quasilinearization, and approximate quasilinearization iterative methods for the complementary Lidstone boundary value problem1.1,1.2. InSection 6, we will show the monotone convergence of Picard’s iterative method. Since the proofs of most of the results in Sections3–6are similar to those of our previous work1,2the details are omitted; however, through some simple examples it is shown how easily these results can be applied in practice.
2. Interpolating Polynomial
Lemma 2.1see47. Lety∈C2m0,1.Then,
yt Q2m−1t Et, 2.1
whereQ2m−1tis the Lidstone interpolating polynomial of degree2m−1,
Q2m−1t m−1
i0
y2i0Λi1−t y2i1Λit
, 2.2
andEtis the residue term
Et 1
0
gmt, sy2msds, 2.3
here
Λ0t t, Λit Λi−1t, Λi0 Λi1 0, i≥1, 2.4
g1t, s ⎧ ⎨ ⎩
t−1s, s≤t, s−1t, t≤s,
git, s 1
0
g1t, t1gi−1t1, sdt1, i≥2.
2.5
Recursively, it follows that
Λit 1
0
git, ssds
1 6
6t2i1 2i1!−
t2i−1 2i−1!
−i−2
k0
222k3−1
2k4! B2k4
t2i−2k−3 2i−2k−3!
22i1 2i1!B2i1
1t
2
, i≥1
2.6
B2i1t is the Bernoulli polynomial of degree 2i1,and B2k4 is the2k 4th Bernoulli
number B2k1 0, k 1,2,3, . . .;B0 1, B1 −1/2, B2 1/6, B4 −1/30, B6 1/42,
Lemma 2.2see47. The following hold:
gmt, s ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ g1
mt, s − m−1
i0
Λit1−s 2m−2i−1
2m−2i−1!, t≤s,
g2
mt, s − m−1
i0
Λi1−t s 2m−2i−1
2m−2i−1!, s≤t,
2.7
0≤−1mgmt, s gmt, s, 2.8
1
0
gmt, sds −1mE2mt≤−1mE2m1
2
−1mE2m
22m2m! 2.9
(E2mt is the Euler polynomial of degree2m,and E2m is the 2mth Euler number E2m1 0, m0,1,2, . . .;E01,E2−1,E45,E6−61)
1
0
gmt, sds −1m2E2mt 1−2tE2m−1t≤−1mE2m−10
−1m12
22m−1 2m! B2m.
2.10
Theorem 2.3. Letx∈C2m10,1.Then,
xt P2mt Rt, 2.11
whereP2mtis the complementary Lidstone interpolating polynomial of degree2m,
P2mt x0
m
i1
x2i−10vi1−vi1−t x2i−11vit−vi0, 2.12
andRtis the residue term
Rt 1
0
hmt, sx2m1sds, 2.13
here
hmt, s t
0
gmτ, sdτ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
−m i1
vit−vi0
1−s2m−2i1
2m−2i1!, t≤s,
s2m 2m!
m
i1
vi1−t−vi1 s
2m−2i1
2m−2i1!, s≤t,
2.14
Λ
Remark 2.4. From2.4and2.15it is clear thatv0t 1;vit Λi−1t,i≥1; 1
0visds0, i≥1;vi0 0,i≥1;vi1 0,i≥2;vit t0vi−1sds,i≥1;
v0t 1, v1t t 2
2 − 1
6, v2t
t4
24−
t2
12 7
360. 2.16
Proof. In2.1, we letyt xtand integrate both sides from 0 tot,to obtain
t
0
xτdτ xt−x0
m−1
i0
x2i10
t
0
Λi1−τdτx2i11 t
0
Λiτdτ
t
0 1
0
gmτ, sx2m1sds
dτ.
2.17
Now, since
t
0
Λiτdτ t
0 Λ
i1τdτ Λi1t−Λi10 vi1t−vi10, i≥0, 2.18
and, similarly
t
0
Λi1−τdτ Λi1 1−Λi11−t vi11−vi11−t, i≥0, 2.19
it follows that
xt x0
m
i1
x2i−10vi1−vi1−t x2i−11vit−vi0
t
0 1
0
gmτ, sx2m1sds
dτ
P2mt Rt.
2.20
Next since
Rt t
0 1
0
gmτ, sx2m1sds
dτ 1
0 t
0
gmτ, sdτ
fort≤s,from2.7, we get
hmt, s t
0
gmτ, sdτ t
0
gm1τ, sdτ
−m−1 i0
t
0
Λiτdτ
1−s2m−2i−1 2m−2i−1!
−m i1
vit−vi01−s
2m−2i1
2m−2i1!, t≤s,
2.22
and similarly, fors≤t,we have
hmt, s t
0
gmτ, sdτ s
0
gm1τ, sdτ t
s
gm2τ, sdτ
−m i1
vis−vi01−s
2m−2i1
2m−2i1!
m
i1
vi1−t−vi1−s s 2m−2i1
2m−2i1!.
2.23
Finally, since2.12is exact for any polynomial of degree up to 2m,we find
t−s2m 2m!
−s2m 2m!
m
i1
−s2m−2i1
2m−2i1!vi1−vi1−t
1−s2m−2i1
2m−2i1!vit−vi0
,
2.24
and hence, forts,it follows that
s2m 2m!
m
i1
s2m−2i1
2m−2i1!vi1−vi1−s−
1−s2m−2i1
2m−2i1!vis−vi0
. 2.25
Combining2.23and2.25, we obtain
hmt, s t
0
gmτ, sdτ s2m 2m!
m
i1
vi1−t−vi1
s2m−2i1
2m−2i1!, s≤t. 2.26
Theorem 2.5. Letx∈C2m10,1.Then, inequalities1.5hold with
C2m1,0 −1m4
22m2−1
2m2! B2m2,
C2m1,2k−1 −
1m−k1E2m−2k2
22m−2k22m−2k2!, 1≤k≤m,
C2m1,2k −1m−k2
22m−2k2−1
2m−2k2!B2m−2k2, 1≤k≤m
C3,0 1/12, C3,1 1/8, C3,2 1/2, C5,0 1/120. C5,1 5/384, C5,2 1/24, C5,3 1/8, C5,41/2.
Proof. From2.14and2.8it follows that
0≤−1mhmt, s |hmt, s|. 2.28
Now, from2.11and2.13, we find
|xt−P2mt| ≤max
0≤t≤1 1
0
|hmt, s|ds
max
0≤t≤1
x2m1t. 2.29
However, from2.9, we have
1
0
|hmt, s|ds 1
0
t
0
gmτ, sdτds t
0 1
0
gmτ, sdsdτt 0
−1mE2mτdτ.
2.30
Thus, from−1mE2mτ ≥0,τ ∈0,1,E2m1τ E2mτ,andE2m10 E2m11 0,we
obtain
1
0
|hmt, s|ds≤ 1
0
−1mE2m1τdτ
−1mE2m11−E2m10 −1m12E2m10
−1m24
22m2−1
2m2! B2m2C2m1,0.
2.31
Using the above estimate in2.29, the inequality1.5fork0 follows. Next, from2.11,2.13and2.14, we have
xjt−P2mjt 1
0
and hence in view of2.5and2.9it follows that
x2k−1t−P2m2k−1t≤max
0≤t≤1 1
0
gm2k−2t, sds
max
0≤t≤1
x2m1t
max
0≤t≤1 1
0
gm−k1t, sdsmax 0≤t≤1
x2m1t
≤ −1m−k1E2m−2k2
22m−2k22m−2k2!max0≤t≤1
x2m1t
C2m1,2k−1max 0≤t≤1
x2m1t, 1≤k≤m,
2.33
and similarly, by2.5and2.10, we get
x2kt−P2m2kt≤max
0≤t≤1 1
0
g2km −1t, sds
max
0≤t≤1
x2m1t
max
0≤t≤1 1
0
gm−k1t, sds
max
0≤t≤1
x2m1t
≤−1m−k2
22m−2k2−1
2m−2k2!B2m−2k2max0≤t≤1
x2m1t
C2m1,2kmax 0≤t≤1
x2m1t, 1≤k≤m.
2.34
Remark 2.6. From2.13,2.28, and the above considerations it is clear that
Rt
1
0
hmt, sds
x2m1ξ E2m1t−E2m10x2m1ξ, 0< ξ <1. 2.35
Remark 2.7. Inequality1.5 with the constants C2m1,k given in2.27is the best possible, as equalities hold for the functionxt E2m1t−E2m10 polynomial of degree2m
1 whose complementary Lidstone interpolating polynomialP2mt ≡ 0,and only for this function up to a constant factor.
Remark 2.8. From the identitysee47, equation1.2.21
∞
k1
1
k2m2 −1
m 2π2m2
22m2!B2m2, 2.36
we have ∞ k1 1 k2 π2 6 ≥
2π2m2
and hence
|B2m2| ≤
π2
3
2m2!
2π2m2. 2.38
We also have the estimatesee47, equation1.2.41
|E2m2| ≤
2
π 2m1
2m2!. 2.39
Thus, from2.27,2.38, and2.39, we obtain
C2m1,0≤ 43π
1
π 2m1
, C2m1,2k−1≤ π
2
1
π
2m−2k2 ,
C2m1,2k≤ 23π
1
π
2m−2k1
, 1≤k≤m.
2.40
Therefore, it follows that
C2m1,k ≤ 4π
3
1
π
2m1−k
, 0≤k≤2m. 2.41
Combining1.5and2.41, we get
xkt−P2mkt≤ 4π
3
1
π
2m1−k
max
0≤t≤1
x2m1t, k0,1, . . . ,2m. 2.42
Hence, ifx∈C∞0,1,for a fixedkasm → ∞,P2mktconverges absolutely and uniformly toxktin0,1,provided that there exists a constantλ,|λ|< π and an integernsuch that
x2m1t Oλ2m1−kfor allm≥n,t∈0,1.
In particular, the functionxt cosλt,t∈0,1satisfies the above conditions. Thus, for each fixedk,expansions
x2kt −1kλ2kcosλt −1kλ2k
1
∞
i1
−1iλ2i−1sinλvit−vi0
, 2.43
x2k1t −1k1λ2k1sinλt −1kλ2k ∞
i1
−1iλ2i−1sinλΛi−1t 2.44
Remark 2.9. Ifx∈C2m1a, b,then
P2mtxa m
i1
b−a2i−1x2i−1avi1−vib−t b−a
x2i−1b
vi t−a
b−a
−vi0
,
2.45
Rt b−a2m b
a hm
t−a b−a,
s−a b−a
x2m1sds. 2.46
Thus, in view of10visds0,i≥1 we have
b
a
P2mtdt b−axa m
i1
b−a2ix2i−1avi1−x2i−1bvi0
. 2.47
Now, sinceBkt kBk−1t,Bk1−t −1kBkt,k1,2, . . . ,from2.6, we find
Λ it
22i
2i!B2i
1t
2
22i
2i!B2i
1−t
2
, 2.48
and hence by2.15it follows that
vi0 Λi0
22i 2i!B2i
1 2
22i
2i!
21−2i−1B 2i,
vi1 Λi1 2
2i
2i!B2i.
2.49
Using these relations in2.47, we obtain an approximate quadrature formula
b
a
xtdt b−axa m
i1
b−a2iB2i
22i 2i!
x2i−1a−21−2i−1x2i−1b. 2.50
2.46we have
e b
a
Rtdt b−a2m2 1
0 1
0
hmt, sx2m1asb−ads
dt
b−a2m2 1
0 1
0
hmt, sds
dt
x2m1ξ, a < ξ < b
b−a2m2 1
0
E2m1t−E2m10dt
x2m1ξ
b−a2m2−E2m10x2m1ξ
2
22m2−1
2m2! B2m2b−a
2m2x2m1ξ.
2.51
Thus, it immediately follows that
|e| b
a
xtdt−b−axa− m
i1
b−a2iB2i
22i 2i!
x2i−1a−21−2i−1x2i−1b
≤−1m2
22m2−1
2m2! B2m2b−a
2m2max t∈a,b
x2m1t.
2.52
From2.52it is clear that2.50is exact for any polynomial of degree at most2m.
Further, in2.52equality holds for the functionxt E2m1t−a/b−a−E2m10and only for this function up to a constant factor.
We will now present two examples to illustrate the importance of2.50and2.52.
Example 2.10. Consider integratingt141over0,1.Here,a0,b1,andxt t141∈
C∞0,1.The exact value of the integral is
1
0
t141dt1 1
15. 2.53
InTable 1, we list the approximates of the integral using2.50with different values ofm,the actual errors incurred, and the error bounds deduced from2.52.
Note thatx15t ≡0,hence the errore 0 when2m1 15 orm 7.Although the errors for other values ofm<7are large, ultimately the approximates tend to the exact value asm → ∞.
Example 2.11. Consider integrating sin 2t over0, π/2. Here,a 0, b π/2,and xt
sin 2t∈C∞0, π/2.The exact value of the integral is
π/2
0
Table1
m Approximate2.50 Actual error|e| Error bound2.52
1 31
3 2
4
15 91
2 −392
15 40
1
5 1001
3 45319
45 452
16
45 7293
4 −31787
9 3179
38
45 31031
5 123215
9 12320
22
45 62881
6 −191114
15 19112
1
3 38227
7 11
15 0 0
Table2
m Approximate2.50 Actual error|e| Error bound2.52
1 0.822467 0.177533 2.029356
2 0.957757 0.042243 2.002894
3 0.989549 0.010451 2.000310
4 0.997394 0.002606 2.000034
5 0.999349 0.000651 2.0000038
6 0.999837 0.000163 2.00000042
7 0.999959 0.000041 2.000000046
InTable 2, we list the approximates of the integral using2.50with different values ofm,the actual errors incurred, and the error bounds deduced from2.52.
Unlike Example 2.10, here the error decreases as mincreases. In both examples, the approximates tend to the exact value asm → ∞.Of course, for increasing accuracy, instead of taking large values ofm,one must use composite form of formula2.50.
3. Existence and Uniqueness
The equalities and inequalities established inSection 2will be used here to provide necessary and sufficient conditions for the existence and uniqueness of solutions of the complementary Lidstone boundary value problem1.1,1.2.
Theorem 3.1. Suppose thatMk>0,0≤k≤qare given real numbers and letQbe the maximum of
|ft, x0, x1, . . . , xq|on the compact set0,1×D0,where
Further, suppose that
QC2m1,k ≤Mk, max
t∈0,1
P2mktpk≤Mk, 0≤k≤q, 3.2
then, the boundary value problem1.1,1.2has a solution inD0.
Proof. The set
B0,1
xt∈Cq0,1:xk max
t∈0,1
xkt≤2Mk, 0≤k≤q
3.3
is a closed convex subset of the Banach spaceCq0,1.We define an operatorT :Cq0,1 →
C2m0,1as follows:
Txt P2mt 1
0
|hmt, s|fs,xsds. 3.4
In view ofTheorem 2.3and2.28it is clear that any fixed point of3.4is a solution of the boundary value problem1.1,1.2. Letxt∈B0,1.Then, from1.5,3.2, and3.4, we find
Txkt≤MkQC2m1,k 2Mk, 0≤k≤q. 3.5
Thus,TB0,1 ⊆ B0,1.Inequalities3.5imply that the sets {Txkt : xt ∈ B0,1}, 0≤k≤qare uniformly bounded and equicontinuous in0,1.Hence,TB0,1that is compact follows from the Ascoli-Arzela theorem. The Schauder fixed point theorem is applicable and a fixed point ofTinD0exists.
Corollary 3.2. Assume that the functionft, x0, x1, . . . , xqon0,1×Rq1satisfies the following
condition:
ft, x0, x1, . . . , xq≤L q
i0
Li|xi|λi, 3.6
whereL, Li,0≤i≤qare nonnegative constants, and0≤λi<1,0≤i≤q,then, the boundary value problem1.1,1.2has a solution.
Theorem 3.3. Suppose that the functionft, x0, x1, . . . , xqon 0,1×D1 satisfies the following
condition:
f
t, x0, x1, . . . , xq≤L q
i0
where
D1
x0, x1, . . . , xq
:|xk| ≤pkC2m1,k Lc
1−θ, 0≤k≤q
, 3.8
c q
i0
Lipi, 3.9
θ q
i0
C2m1,iLi<1, 3.10
then, the boundary value problem1.1,1.2has a solution inD1.
Theorem 3.4. Suppose that the differential equation1.1together with the homogeneous boundary
conditions
x0 0, x2i−10 0, x2i−11 0, 1≤i≤m 3.11
has a nontrivial solutionxtand the condition3.7withL0is satisfied on0,1×D2,where
D2x0, x1, . . . , xq:|xk| ≤C2m1,kM, 0≤k≤q 3.12
andMmaxt∈0,1|x2m1t|,then, it is necessary thatθ≥1.
Remark 3.5. Conditions ofTheorem 3.4ensure that in3.7at least one of theLi, 0≤i≤qwill
not be zero; otherwise the solutionxtwill be a polynomial of degree at most 2mand will not be a nontrivial solution of1.1,1.2. Further,xt ≡ 0 is obviously a solution of1.1,
1.2, and ifθ <1,then it is also unique.
Theorem 3.6. Suppose that for allt, x0, x1, . . . , xq,t, x0, x1, . . . , xq∈0,1×D1the functionf
satisfies the Lipschitz condition
f
t, x0, x1, . . . , xq−ft, x0, x1, . . . , xq≤ q
i0
Li|xi−xi|, 3.13
where L maxt∈0,1|ft,0,0, . . . ,0|,then, the boundary value problem1.1,1.2 has a unique
solution inD1.
Example 3.7. Consider the complementary Lidstone boundary value problem
−x3t ft, x, x, . . . , xq, t∈0,1, 3.14
where 0 ≤ q ≤ 2 is fixed. Here, m 1 and the interpolating polynomial satisfying1.4is computed asP2t 1−tt2with
p0max
t∈0,1|P2t|P20 1, p1tmax∈0,1P
2tP21 1, p2 max t∈0,1P
2t2. 3.16
We illustrateTheorem 3.1by the following two cases.
Case 1. Suppose q 0 and ft, x tx2, then,Theorem 3.1states that 3.14, 3.15has a
solution in the setD0{x:|x| ≤2M0}provided
M0≥p01, QC3,0≤M0. 3.17
We will look for a constantM0that satisfies3.17. Since
Q max
t,x∈0,1×D0
ft, x 2M02, 3.18
the conditionQC3,0 ≤M0simplifies to 0≤M0≤3.Coupled with another conditionM0 ≥1,
we see that 1≤M0≤3 fulfills3.17. Therefore, we conclude that the differential equation
−x3t tx2, t∈0,1 3.19
with the boundary conditions3.15has a solution inD0{x:|x| ≤2M0}whereM0∈1,3.
Case 2. Supposeq2 andft, x, x, x t2x√tx t/2x,then,Theorem 3.1states that
3.14,3.15has a solution in the setD0 {x, x, x: |x| ≤ 2M0,|x| ≤ 2M1,|x| ≤2M2}
provided
Mk≥pk, QC3,k ≤Mk, k0,1,2. 3.20
Here
Q max
t,x,x,x∈0,1×D
0
ft, x, x, x2M02M1M2, 3.21
and the conditionsQC3,k≤Mk,k0,1,2,reduce to
10M0−2M1−M2≥0, −2M06M1−M2≥0, −2M0−2M1M2≥0. 3.22
PickM0 1,M1 1,M2 4 which satisfy3.22and alsoMk ≥ pk,k 0,1,2.It follows
fromTheorem 3.1that the differential equation
−x3t t2x√txt
2
with the boundary conditions3.15 has a solution in D0 {x, x, x : |x| ≤ 2, |x| ≤ 2,
|x| ≤8}.
Example 3.8. Consider the complementary Lidstone boundary value problem
−x3t sint sintx costxx
4 , t∈0,1 3.24
with the boundary conditions3.15. Here,m 1,q 2 and the interpolating polynomial
P2tsatisfying1.4is given inExample 3.7. To illustrateTheorem 3.3, we note that fort ∈ 0,1and anyx0, x1, x2,
ft, x0, x1, x2sint sintx0 costx1 x2
4
≤1|x0||x1||x2|
4 . 3.25
Thus, condition3.7is satisfied withL 1,L0 1,L1 1,L2 1/4.The constantscandθ
are then computed as
c 2
i0
Lipi 5
2, θ
2
i0
C3,iLi 1
3 <1. 3.26
ByTheorem 3.3, problem3.24,3.15has a solution in
D1 x, x, x:|x| ≤ 23 16, x
≤ 53
32, x
≤ 37
8
. 3.27
4. Picard’s and Approximate Picard’s Methods
Picard’s method of successive approximations has an important characteristic, namely, it is constructive; moreover, bounds of the difference between iterates and the solution are easily available. In this section, we will provide a priori as well as posteriori estimates on the Lipschitz constants so that Picard’s iterative sequence {xnt} converges to the unique solutionx∗tof the problem1.1,1.2.
Definition 4.1. A functionxt∈C2m10,1is called anapproximate solutionof1.1,1.2if there exist nonnegative constantsδandsuch that
max
t∈0,1
−1mx2m1t−ft,xt≤δ, 4.1
max
t∈0,1
whereP2mtandP2mtare polynomials of degree 2msatisfying1.2, and
P0 x0, P2i−10 x2i−10, P2i−11 x2i−11, 0≤i≤m, 4.3
respectively.
Inequality4.1means that there exists a continuous functionηtsuch that
−1mx2m1t ft,xt ηt,
max
t∈0,1ηt≤δ.
4.4
Thus, fromTheorem 2.3the approximate solutionxtcan be expressed as
xt P2mt 1
0
|hmt, s|
fs,xs ηsds. 4.5
In what follows, we will consider the Banach spaceBCq0,1and forx∈Cq0,1,
xmax
0≤k≤q C
2m1,0 C2m1,ktmax∈0,1
xkt
. 4.6
Theorem 4.2. With respect to the boundary value problem1.1,1.2one assumes that there exists
an approximate solutionxt,and
ithe functionft, x0, x1, . . . , xqsatisfies the Lipschitz condition3.13on0,1×D3,where
D3x0, x1, . . . , xq:xk−xkt≤NC2m1,k
C2m1,0, 0≤k≤q, N >0
, 4.7
iiN0 1−θ−1δC2m1,0≤N.
Then, the following hold:
1there exists a solutionx∗tof1.1,1.2inSx, N0 {x∈B:x−x ≤N0},
2x∗tis the unique solution of1.1,1.2inSx, N, 3the Picard iterative sequence{xnt},defined by
xn1t P2mt 1
0
|hmt, s|fs,xnsds, n0,1, . . . , 4.8
wherex0t xtconverges tox∗twithx∗−xn ≤θnN0,and
x∗−xn ≤θ1−θ−1xn−xn
−1, 4.9
InTheorem 4.2conclusion3ensures that the sequence{xnt}obtained from4.8
converges to the solution x∗t of the boundary value problem 1.1, 1.2. However, in practical evaluation this sequence is approximated by the computed sequence, say,{znt}.To
findzn1t,the functionfis approximated byfn.Therefore, the computed sequence{znt}
satisfies the recurrence relation
zn1t P2mt 1
0
|hmt, s|fns,znsds, n0,1, . . . , 4.10
wherez0t x0t xt.
With respect tofnwe will assume the following condition.
Condition C1. Forzntobtained from4.10, the following inequality holds:
ft,znt−fnt,znt≤μft,znt, n0,1, . . ., 4.11
whereμis a nonnegative constant.
Inequality4.11corresponds to the relative error in approximating the functionfby
fnfor then1th iteration.
Theorem 4.3. With respect to the boundary value problem1.1,1.2one assumes that there exists
an approximate solutionxt,and Condition C1is satisfied. Further, one assumes that
icondition (i) ofTheorem 4.2,
iiθ1 1μθ <1,
iiiN1 1−θ1−1δμFC2m1,0≤N,whereFmaxt∈0,1|Ft,xt|,
then,
1all the conclusions (1)–(4) ofTheorem 4.2hold,
2the sequence{znt}obtained from4.10remains inSx, N1,
3the sequence {znt} converges to x∗t, the solution of 1.1, 1.2 if and only if limn→ ∞an0,where
an
zn1t−P2mt− 1
0
|hmt, s|fs,znsds
, 4.12
and the following error estimate holds
x∗−zn1 ≤1−θ−1
θzn1−znμC2m1,0max
t∈0,1ft,znt
. 4.13
Condition C2. Forzntobtained from4.10, the following inequality is satisfied:
ft,znt−fnt,znt≤ν, n0,1, . . . , 4.14
whereνis a nonnegative constant.
Inequality4.14corresponds to the absolute error in approximating the functionfby
fnfor then1th iteration.
Theorem 4.4. With respect to the boundary value problem1.1,1.2one assumes that there exists
an approximate solutionxt,and Condition C2is satisfied. Further, one assumes that
icondition (i) ofTheorem 4.2,
iiN2 1−θ−1δνC2m1,0≤N,
then,
1all the conclusions (1)–(4) ofTheorem 4.2hold,
2the sequence{znt}obtained from4.10remains inSx, N2,
3the sequence {znt} converges to x∗t, the solution of 1.1, 1.2 if and only if limn→ ∞an0,and the following error estimate holds:
x∗−zn1 ≤1−θ−1θz
n1−znνC2m1,0. 4.15
Example 4.5. Consider the complementary Lidstone boundary value problem
−x3t 1xxx
4 , t∈0,1 4.16
with the boundary conditions3.15. PickP2t 1−tt2to be an approximate solution of 4.16,3.15, that is, letxt P2t.Then, from4.2we get 0.Further, from4.1we have
max
t∈0,1
−x3t−ft, xt, xt, xt
max
t∈0,1
ft, xt, xt, xt
max
t∈0,1
1xt xt x
t
4
max
t∈0,1
32 tt2 7
2 ≡δ.
To illustrateTheorem 4.2, we note that the Lipschitz condition3.13is satisfiedgloballywith
L01,L11,L21/4,and the constantsθandN0are computed directly as
θ 2
i0
C3,iLi 1
3, N0 1−θ
−1δC3,0 21
4 ≤N. 4.18
ByTheorem 4.2, it follows that
1there exists a solutionx∗tof4.16,3.15inSP2, N0, 2x∗tis the unique solution of4.16,3.15inSP2, N, 3the Picard iterative sequence{xnt}defined by
−x3n1t 1xnt xnt
xnt
4 , n0,1, . . . ,
xn10 1, xn1 0 −1, xn11 1,
4.19
wherex0t P2tconverges tox∗twith
x∗−xn ≤1
3
n21
4 , x
∗−xn ≤ 1
2xn−xn−1. 4.20
Suppose that we require the accuracyx∗−xn ≤10−5,then from above we just set
1 3
n21
4 ≤10
−5 4.21
to getn≥12.Thus,x12twill fulfill the required accuracy.
Finally, we will illustrate how to obtainx1tfrom4.19. First, we integrate
−x31 t 1x0t x0t x 0t
4 3 2 tt
2 4.22
from 0 totto get
−x
1t x10
3t
2
t2
2
t3
3. 4.23
Next, integrating4.23from 0 totas well as fromtto 1,respectively, gives
−x
1t x10 tx10
3t2
4
t3
6
t4
12, 4.24
−x
11 x1t 1−tx10 1−
3t2
4 −
t3
6 −
t4
Adding4.24and4.25yieldsx10 3.Now, integrate4.24 or4.25from 0 totgives
x1t 1−t3t 2
2 −
t3
4 −
t4
24−
t5
60. 4.26
A similar method can be used to obtainxnt,n≥2.
5. Quasilinearization and Approximate Quasilinearization
Newton’s method when applied to differential equations has been labeled as quasilineariza-tion. This quasilinear iterative scheme for1.1,1.2is defined as
−1mx2m1n1 t ft,xnt βt q
i0
xin1t−xin t ∂
∂xin t
ft,xnt, 5.1
xn10 α0, xn12i−10 αi, xn12i−11 βi, 0≤i≤m, n0,1, . . . , 5.2
wherex0t xtis an approximate solution of1.1,1.2.
In the following results once again we will consider the Banach spaceCq0,1and for
x∈Cq0,1the normxis as in4.6.
Theorem 5.1. With respect to the boundary value problem1.1,1.2one assumes that there exists
an approximate solutionxt,and
ithe functionft, x0, x1, . . . , xqis continuously differentiable with respect to allxi,0≤i≤ qon0,1×D3,
iithere exist nonnegative constantsLi,0≤i≤qsuch that for allt, x0, x1, . . . , xq∈0,1× D3,
∂xi∂ ft, x0, x1, . . . , xq≤Li, 5.3
iiithe functionβtis continuous on0,1,βmaxt∈0,1|βt|,andθβ 12βθ <1, ivN3 1−θβ−1δC2m1,0≤N.
Then, the following hold:
1the sequence{xnt}generated by the iterative scheme5.1,5.2remains inSx, N3, 2the sequence{xnt}converges to the unique solutionx∗tof the boundary value problem
1.1,1.2,
3a bound on the error is given by
xn−x∗ ≤
1βθ
1−βθ n
Theorem 5.2. Let in Theorem 5.1 the functionβt ≡ 1.Further, letft, x0, x1, . . . , xq be twice continuously differentiable with respect to allxi,0≤i≤qon0,1×D3,and
∂2 ∂xi∂xjf
t, x0, x1, . . . , xq
≤LiLjK, 0≤i, j≤q. 5.5
Then,
xn1−xn ≤αxn−xn−12≤
1
ααx1−x0 2n
≤ 1 α
1
2Kδ
θ
1−θ 2!2n
, 5.6
whereαKθ2/21−θC
2m1,0.Thus, the convergence is quadratic if
1
2Kδ
θ
1−θ 2
<1. 5.7
Conclusion3ofTheorem 5.1ensures that the sequence{xnt}generated from the scheme 5.1,5.2 converges linearly to the unique solution x∗t of the boundary value problem1.1,1.2.Theorem 5.2provides sufficient conditions for its quadratic convergence. However, in practical evaluation this sequence is approximated by the computed sequence, say,{znt}which satisfies the recurrence relation
−1mz2m1n1 t fnt,znt βt q
i0
zin1t−zin t ∂
∂zin t
fnt,znt,
zn10 α0, z2in1−10 αi, zn12i−11 βi, 0≤i≤m, n0,1, . . . ,
5.8
wherez0t x0t xt.
With respect tofnwe will assume the following condition.
Condition C3. fnt, x0, x1, . . . , xqis continuously differentiable with respect to allxi, 0≤i≤q
on0,1×D3with
∂xi∂ fnt, x0, x1, . . . , xq≤Li 5.9
and Condition C1is satisfied.
Theorem 5.3. With respect to the boundary value problem1.1,1.2one assumes that there exists
an approximate solutionxt,and the Condition C3is satisfied. Further, one assumes
iconditions (i) and (ii) ofTheorem 5.1,
iiθβ,μ 12βμθ <1,
then,
1all conclusions (1)–(3) ofTheorem 5.1hold,
2the sequence{znt}generated by the iterative scheme5.8, remains inSx, N4, 3the sequence{znt}converges tox∗t,the unique solution of 1.1,1.2if and only if
limn→ ∞an0,and the following error estimate holds:
x∗−zn1 ≤1−θ−11βθzn1−znμC2m1,0max
t∈0,1ft,znt
. 5.10
Theorem 5.4. Let the conditions ofTheorem 5.3be satisfied. Further, letfn f0for alln1,2, . . .
and f0t, x0, x1, . . . , xq be twice continuously differentiable with respect to allxi, 0 ≤ i ≤ q on 0,1×D3,and
∂2 ∂xi∂xjf0
t, x0, x1, . . . , xq
≤LiLjK, 0≤i, j ≤q. 5.11
Then,
zn1−zn ≤αzn−zn−12≤
1
ααz1−z0 2n
≤ 1 α
1
2KδμF
θ
1−θ 22n
, 5.12
whereαis the same as inTheorem 5.2.
Example 5.5. Consider the complementary Lidstone boundary value problem
−x3t tx2, t∈0,1 5.13
again with the boundary conditions3.15. First, we will illustrateTheorem 5.1. Pickxt 0 andβt 1soβ1. Clearly,ft, x tx2is continuously differentiable with respect tox
for allt, x.Forx∈D3{x:|x| ≤N},we have
∂x∂ ft, x|2x| ≤2N≡L0. 5.14
Thus,
θC3,0L0 N
6 , θβ
12βθ N
2 . 5.15
LetN <2 so thatθβ <1.Next, from4.1we have maxt∈0,1|ft,0|1≡δ.Also, from4.2
we find
max
t∈0,1
P2t−P2t max
t∈0,1|P2t|1≤C3,0
and so we take12.Now,
N3
1−θβ −1
δC3,0
13
6N ≤N 5.17
yieldsN≥"13/6 1.633.Coupled withN <2so thatθβ<1, we should impose
#
13
6 ≤N <2. 5.18
The corresponding range ofN3will then be
13
12 < N3≤
#
13
6 . 5.19
The conditions ofTheorem 5.1are satisfied and so
1the sequence{xnt}generated by
−x3n1t tx2
nt 2xn1t−xntxnt, n0,1, . . . , xn10 1, xn1 0 −1, xn11 1,
5.20
wherex0t 0 remains inS0, N3,that is, maxt∈0,1|xnt| ≤N3,
2the sequence{xnt}converges to the unique solutionx∗tof5.13,3.15with
max
t∈0,1|x
∗t−xnt| ≤ 2N
6−N
n 13
62−N. 5.21
Next, we will illustrateTheorem 5.2. Forx∈D3{x:|x| ≤N},we have
∂2 ∂x2ft, x
2≤L20K 2N2K. 5.22
Hence, we may takeK1/2N2.FromTheorem 5.2, we have
max
t∈0,1|xn1t−xnt| ≤
1
α
1
2Kδ
θ
1−θ 22n
26−N
13 46−N2
2n
. 5.23
The convergence is quadratic if
1
2Kδ
θ
1−θ 2
which is the same as
13
4 <6−N
2 5.25
and is satisfied if N > 7.803 or N < 4.197.Combining with5.18, we conclude that the convergence of the scheme5.20is quadratic if
#
13
6 ≤N <2. 5.26
6. Monotone Convergence
It is well recognized that the method of upper and lower solutions, together with uniformly monotone convergent technique offers effective tools in proving and constructing multiple solutions of nonlinear problems. The upper and lower solutions generate an interval in a suitable partially ordered space, and serve as upper and lower bounds for solutions which can be improved by uniformly monotone convergent iterative procedures. Obviously, from the computational point of view monotone convergence has superiority over ordinary convergence. We will discuss this fruitful technique for the boundary value problem1.1,
1.2withq1.
Definition 6.1. A functionμt ∈C2m10,1is said to be alower solutionof1.1,1.2with
q1 provided
−1mμ2m1t≤ft, μt, μt, t∈0,1,
μ0−α0
≤0, −1i−1μ2i−10−αi
≤0, −1i−1μ2i−11−βi
≤0, 1≤i≤m. 6.1
Similarly, a functionνt∈C2m10,1is said to be anupper solutionof1.1,1.2withq1 if
−1mν2m1t≥ft, νt, νt, t∈0,1,
ν0−α0≥0, −1i−1ν2i−10−αi≥0, −1i−1ν2i−11−βi≥0, 1≤i≤m. 6.2
Lemma 6.2. Letμtandνtbe lower and upper solutions of1.1,1.2withq1,and letP2m,μt
andP2m,νtbe the polynomials of degree2msatisfying
and
P2m,ν0 μ0, P2m,ν2i−10 ν2i−10, P2m,ν2i−11 ν2i−11, 1≤i≤m, 6.4
respectively. Then, for allt∈0,1,P2m,μk t≤P2mkt≤P2m,νk t,k0,1.
Proof. From2.5,2.6, and2.8it is clear that−1iΛit≥0,−1iΛi1−t≥0,i≥0 and this
in turn from2.18and2.19implies that−1ivi1t−vi10≥0,−1ivi11−vi11−t≥
0,−1ivi1t −1iΛit≥0,−1ivi11−t −1iΛi1−t≥0,i≥0.Now, since
P2m,μt μ0 m
i1
μ2i−10vi1−vi1−t μ2i−11vit−vi0
,
P2m,μ t m
i1
μ2i−10Λi−11−t μ2i−11Λi−1t
,
6.5
it follows that
P2m,μt μ0 m
i1
−1i−1μ2i−10−1i−1vi1−vi1−t
−1i−1μ2i−11−1i−1vit−vi0
≤α0 m
i1
−1i−1αi−1i−1vi1−vi1−t −1i−1βi−1i−1vit−vi0
P2mt.
6.6
Similarly, we haveP2m,μ t≤P2m t.The proof ofP2mkt≤P2m,νk t, k0,1 is similar.
In the following result for xt ∈ C10,1 we will consider the norm x
max{maxt∈0,1|xt|,maxt∈0,1|xt|}and introduce a partial orderingas follows. Forx, y∈ C10,1we say thatxyif and only ifxt≤ytandxt≤ytfor allt∈0,1.
Theorem 6.3. With respect to the boundary value problem 1.1,1.2 with q 1 one assumes
that ft, x0, y0 is nondecreasing in x0 and y0.Further, let there exist lower and upper solutions μ0t, ν0tsuch thatμ0ν0.Then, the sequences{μnt},{νnt}whereμntandνntare defined
by the iterative schemes
μn1t P2mt 1
0
gmt, sf
s, μns, μnsds, n0,1, . . . ,
νn1t P2mt 1
0
gmt, sf
s, νns, νns
ds, n0,1, . . .
are well defined, and{μnt}converges to an elementμt∈C10,1,{νnt}converges to an element νt∈C10,1(with the convergence being in the norm ofC10,1). Further,μ
0μ1 · · · μn · · · μν · · · νn · · · ν1 ν0,μt, νtare solutions of 1.1,1.2withq1,and each
solutionztof this problem which is such thatz∈μ0, ν0satisfiesμzν.
Example 6.4. Consider the complementary Lidstone boundary value problem
−x3t 1xx, t∈0,1,
x0 1, x0 −1, x1 −1.
6.8
Here,m1, q1 and the functionft, x0, y0 1x0y0is nondecreasing inx0andy0.We
find that6.8has a lower solution
μ0t 1−t 6.9
and an upper solution
ν0t 18t2− 17
3 t
3 6.10
such that
μ0t≤ν0t, μ0t≤v0t, t∈0,1. 6.11
Hence,μ0 ν0and the conditions ofTheorem 6.3are satisfied. The iterative schemes
−μ3n1t 1μnμn, n0,1, . . . ,
μn10 1, μn10 −1, μn11 −1,
6.12
−νn13t 1νnνn, n0,1, . . . ,
νn10 1, νn10 −1, νn1 1 −1
6.13
will converge respectively to someμ∈C10,1andν∈C10,1.Moreover,
andμt, νtare solutions of6.8. Any solutionztof6.8which is such thatz ∈μ0, ν0
fulfillsμzν.As an illustration, by direct computationas inExample 4.5, we find
μ1t 1−t t2
6 −
t3
6
t4
24,
μ2t 1−t−
29t2
160
t3
6 −
t4
36−
t5
180
t7
5040,
. . . ,
ν1t 1−t−
79t2
60
t3
3 2t4
3 − 3t5
20 − 17t6
360,
ν2t 1−t−
83t2
40320
t3
6 − 109t4
720 − 19t5
3600
t6
40−
t7
2520− 13t8
10080 − 17t9
181440.
. . . .
6.15
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